Let , where 3 and q are relatively prime, and q is odd
Then (1)
Clearly is divisible by if , and is divisible by if.
Did I miss something?

Yes, I did. In the expression (1) figure in brackets is not integer (prove or disprove).
Statement about s=1 is still correct if we move q inside the brackets - there are enough 3s there to compensate, but statement about q still need to be proved.

Let's take another way.
If n = p, p - odd prime (n is odd, obviously), then:, so if n - prime it is 3.

Let , where q - smallest prime, divided n.
Then exists smallest r:. It is true, because q is a divisor of n.
Let , then:
If a - even: r - is not the smallest - contradiction
if a - odd: : - for some m and : - r is not the smallest.
So r divides n. Let's show that :
If : , then , so , and r is not smallest again.
So n cannot be composite and n = 3 is the only solution.

Is it messy?
Personally, I do not like this solution. Can somebody come up with the more elegant one?